TITLE: How many resulting regions if we partition $\mathbb{R}^m$ with $n$ hyperplanes?
QUESTION [15 upvotes]: This is a generalization of this question. So in $\mathbb{R}^2$, the problem is illustrated like so:

Here, $n = 3$ lines divides $\mathbb{R}^2$ into $N_2=7$ regions. For general $n$ in the case of $\mathbb{R}^2$, the number of regions $N_2$ is $\binom{n+1}{2}+1$. But what about if we consider the case of $\mathbb{R}^m$, partitioned using $n$ hyperplanes?
Is the answer $N_m$ still $\binom{n+1}{2}+1$, or will it be a function of $m$?

REPLY [6 votes]: Denote this number as $A(m, n)$. We will prove $A(m, n) = A(m, n-1) + A(m-1, n-1)$. 
Consider removing one of the hyperplanes, the maximum number is $A(m, n-1)$. Then, we add the hyperplane back. The number of regions on the hyperplane is the same as the number of newly-added regions. Since this hyperplane is $m-1$ dimensional, this maximum number is $A(m-1, n-1)$.
So, this number satisfies $A(m, n) = A(m, n-1) + A(m-1, n-1)$, and the formula can be simply derived by induction as $\sum_{i=0}^m \binom{n}{i}$.